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Work Required

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the work required to move a charge of -0.51x10^-12 C from `i' to `b'.
    http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype54/prob04a_threeqcontour.gif

    2. Relevant equations
    i know that the equation is -W=qEl and that it can also be expressed as -W=-q[tex]\Delta[/tex]V or W=q[tex]\Delta[/tex]V


    3. The attempt at a solution
    well q is given as -0.51x10^-12
    the problem i am having is how do i find [tex]\Delta[/tex]V without knowing E? cause i know i can figure out l just by using the picture. So if you can just help me with find out how to get [tex]\Delta[/tex]V that would be great!
     
  2. jcsd
  3. Feb 8, 2009 #2

    LowlyPion

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    Looks like your ΔV is going from +3 to -3. That looks like a Voltage drop of 6v from the image, or ΔV = -6.

    Just count the contours is my method.
     
  4. Feb 8, 2009 #3

    Redbelly98

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    The figure is a contour plot of the potential. Each contour line represents a constant potential value.

    The figure also indicates which contour lines correspond to -5V, 0V, and +5V. Using that, can you say:
    What is the voltage at point i?
    What is the voltage at point b?
     
  5. Feb 9, 2009 #4
    ok so i did the (-.51x10^-12)(-6) and got an answer of 3.06x10^-12 J and it was said to be wrong. Where did i got wrong?
     
  6. Feb 9, 2009 #5
    oh and i was given this hint when i put my answer in

    The equipotential lines shown are separated by 1 kV. (NOTE! That's kV, not volts!) Work to move a charge is the increase in potential energy of the charge. The potential energy is the potential times the charge.
     
  7. Feb 9, 2009 #6

    cjl

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    The answer remains the same though, whether it is kV or V. It's still a matter of counting contour lines.
     
  8. Feb 9, 2009 #7

    Redbelly98

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    If it's actually 1 kV, and not 1 V, for the contour intervals, then instead of 6V it is really ____?
     
  9. Feb 9, 2009 #8
    ok so then it would mean that it was really 6 kV or 6000 V and then when i did that i got 3x10^-9 J and that also came up wrong
     
  10. Feb 9, 2009 #9
    your graph is ill-made....
     
  11. Feb 9, 2009 #10
    haha, it is the graph that they gave up. i didnt make it
     
  12. Feb 9, 2009 #11
    ok i give up... i am having my own probelms...
     
  13. Feb 9, 2009 #12

    LowlyPion

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    It's plus Work moving an opposite charge away from + and toward a - V.

    If the contours are 103v then it's 3.06 * 10-9 J or 306 ergs.

    If it's not precision or units then I've no idea what is asked.
     
  14. Feb 10, 2009 #13
    i got the answer. i did the 3.06e-09 J and it seemed to work even though i did 3.1e-09 before. i guess it had to be 06. thanks for all your help guys
     
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